Limit of $F [v_j]$ for $F [w] := \int_Ω f (x, w(x)) dx$. I was trying to understand a part from a book on calculus of variation, but in the following text, I don't understand how the two problems are equivalent :

Assume that we are given a weakly converging sequence $v_j$  in
$L^2(Ω)$, where $Ω ⊂ \mathbb{R}^d$ is a bounded Lipschitz domain, and
an integral functional $$F [w] := \int_Ω f (x, w(x)) dx, w ∈ L^2
> (Ω)$$, with $f : Ω × \mathbb{R} → \mathbb{R}$ continuous and bounded
(for simplicity). Then, $(F [v_j])_j$ is a bounded sequence and up to
a (non-renumbered) subsequence we may assume that $F [v_j]$ converges
to some limit as $j → ∞$. The question then arises: how can we compute
this limit for every integrand $f$ as above? Equivalently, we could
ask for the weak* limit in $L^∞ (Ω)$ of the sequence of compound
functions $F_j (x) := f (x, v_j (x))$.

Why do we have

Equivalently, we could ask for the weak* limit in $L^∞ (Ω)$ of the
sequence of compound functions $F_j (x) := f (x, v_j (x))$.

?
 A: Let's first unpack what it means for $F[v_j]$ to converge. Suppose $$F[v_j]\rightarrow F^*$$ and note that since f and $\Omega$ are bounded, we must have that
$$\lim_{j\rightarrow \infty}F[v_j]=\lim_{j\rightarrow \infty}\int_\Omega f(x, v_j(x))dx=\int_\Omega \lim_{j\rightarrow \infty} f(x, v_j(x))dx \,\,\,(\text{By Dominated Convergence}).$$ Now note that f is assumed continuous therefore, $\lim_{j\rightarrow\infty} f(x, v_j(x))$ converges pointwise to some measurable function- call it $F(x)$, which yields that $$F^*=\int_\Omega F(x)dx.$$
Now let's look at the definition of weak* convergence in $L^\infty(\Omega)$. We say that a sequence$F_j(x)=f(x, v_j(x))$ converges in the weak* topology to a limit $F(x)$ if and only if $$\int_\Omega F_j(x)g(x)dx\rightarrow \int_\Omega F(x) g(x)dx \,\, , \forall g\in L^1(\Omega).$$
Therefore, the two notions will be equivalent if the following equivalence holds: $$\int_\Omega F_j(x)g(x)dx\rightarrow \int_\Omega F(x) g(x)dx \,\, , \forall g\in L^1(\Omega) \iff \int_\Omega F_j(x)dx \rightarrow \int_\Omega F(x)dx.$$
The $\Rightarrow$ direction is obvious since if the convergence holds for any $g\in L^1(\Omega)$, than just take $g=1$. Conversely, the other direction holds because convergence of the integrals implies we can extract a pointwise convergent sub-sequence such that $F_{jk}[x]\rightarrow F[x]$ pointwise, and then we are free to pull the limit inside of $\int_\Omega F_{jk}(x)g(x)dx$ and the result follows.

Note that this would fail if we got rid of either the continuity or boundedness assumptions on f.
